Saturday, May 29, 2021

H C Verma solutions, ELECTRIC CURRENT IN CONDUCTORS, Exercises, Q41-Q50, Chapter-32, Concepts of Physics, Part-II

Electric Current in Conductors


Exercises, Q41 - Q50


   41.  Find the equivalent resistance of the circuits shown in figure (32-E17) between points a and b. Each resistor has a resistance r.    
Figure for Q-41


ANSWER: The figure in (a) may be rearranged as below, 
The diagram for Q-41(a)

   The upper four resistors 1 to 5 are arranged in a balanced Wheatstone bridge, hence the current will never pass through resistor 5. The resistor 5 is not working. Hence now the equivalent resistance of 1 and 2 =2r, and that

of 3 and 4 =2r. Thus the resistances 2r, 2r and r are in parallel and their equivalent resistance R between a and b is given as

 1/R =1/2r +1/2r +1/r

→1/R =(1+1+2)/2r

→1/R =4/2r =2/r

→R =r/2.


     For figure (b), let us name the resistors and assume that a current i enters point a and the same current i exits at b as shown in the figure below:-
Naming of resistors
Considering the symmetry at points a and b, we assume that the current i at a divides into i₁, i₂ and i₁ in resistors 1, 3 and 2 respectively. Similarly current in resistors 10, 12 and 11 will be i₁, i₂ and i₁ respectively. Now the current in resistor 1 divides as i₃ and i₄ into resistors 4 and 6 respectively. So 

i₃+i₄ =i₁.

Since the current in resistor 10 =i₁, the current in resistor 5 = i₁-i₄ =i₃, away from center O. We can assume that the current i₃ coming towards O in resistor 4  enters in resistor 5 as if their junction is not connected at O. Similarly we can assume that a current i₃ in resistor 7 coming towards O enters the resistor 8 and move away from O as if the junction of resistors 7 and 8 is not connected at O.  And the current i₂ in resistor 3 goes straight through the resistor 12. Now a simplified diagram can be drawn as below:-
Simplified diagram

Equivalent resistance R' of resistances 4, 5 and 6,  

1/R' =1/r +1/2r =3/2r  

→R' =2r/3,

Similarly equivalent resistance of resistances 7, 8 and 9 =R' =2r/3.

Equivalent resistance R" of resistance 3 and 12 (in series) =R" =2r.

Now the resistances are arranged in three rows between a and b. In the upper and lower rows, the resistances in series are r, R' and r. Its equivalent resistance R* =r+R'+r =2r+ 2r/3

→R* =8r/3.

In the middle row, there is resistance R" =2r.

So between and b, resistances R*, R" and R* are in parallel. Their equivalent resistance R is given as,  

1/R =1/R* +1/R" +1/R* 

→1/R =2/R* +1/R"

→1/R =2/(8r/3) +1/2r

→1/R =3/4r +1/2r =5/4r

→R =4r/5.              

           

  


 

   42.  Find the current measured by the ammeter in the circuit shown in figure (32-E18).  
Figure for Q-42



ANSWER: First, let us name the junctions and resistors as below,
Naming Junctions and resistors
  Suppose the current measured by the ammeter =i. So current i enters at junction A and exits at junction D. From symmetry the current in each of the resistors 1, 2, 5 and 6 will be the same, say i₁. This will make potentials at B and F the same = Va-i₁*10. So there is no potential difference between B and F, thus no current in resistor 7.

    Similarly, the potentials at points C and E will be the same = Vd+i₁*10. Hence no potential difference across resistor 8 and no current through it. 

     Thus we conclude that the current through resistors 3 and 4 will be i₁ each.  Now the given combination may be drawn without resistors 7 and 8. This will make the arrangement of resistors as if 1, 3 and 5 in series, with equivalent resistance =10 Ω +10 Ω +10 Ω =30 Ω. Similarly the resistors 2, 4 and 6 in series with equivalent resistance =30 Ω.

    Finally, the arrangement between A and  D becomes two resistances each of 30 Ω in parallel. Thus the equivalent resistance of the whole arrangement R is given as,

1/R = 1/30 +1/30 =2/30 =1/15

→R =15 Ω.

The potential difference across A and D =6 V, hence the reading in ammeter will be

i =6/15 A =0·4 A. 

          

  



 

   43.  Consider the circuit shown in figure (32-E19a). Find (a) the current in the circuit. (b) the potential drop across the 5 Ω resistor, (c) the potential drop across the 10 Ω resistor. (a) Answer the parts (a), (b) and (c) with reference to figure (32-E19b).   
The figure for Q-43



ANSWER: Figure (a) 

   (a) Assume the current in the circuit anticlockwise = i. Apply Kirchhoff's loop law anticlockwise.

-6 +i*5 -12 +i*10 = 0

→15i = 18 

→ i = 18 /15 A =6/5 A =1·2 A. 


(b) The potential drop across 5 Ω resistor = i*5 =1·2*5 V =6 V. 


(c) The potential drop across 10 Ω resistor = i*10 =1·2*10 V =12 V. 


Figure (b)

    (a) We can solve it like figure (a) but it can also be solved as below. 

 The cells are connected in series, the total potential difference across the cells =12+6 =18 V. The 10 Ω and 5 Ω resistors are connected in series, hence their equivalent resistance R =10+5 =15 Ω.

  Since the cells provide a potential difference of 18 V across the resistors, the current through them (current in the circuit),

i =18/15 A =1·2 A. 


(b) The potential drop across 5 Ω resistor =5*i =5*1·2 V = 6 V. 


(c) The potential drop across 10 Ω resistor =10*i =10*1·2 V = 12 V.                  

  




  

   44.  Twelve wires, each having equal resistance r, are joined to form a cube as shown in figure (32-E20). Find the equivalent resistance between the diagonally opposite points a and f.   
The figure for Q -44


ANSWER: Let the equivalent resistance between the diagonally opposite points a and f = R. Suppose a current I enters at point a of the cube and the same current exits at f. From symmetry, the current at the junction a will be equally distributed in three connected resistors. Suppose in each of the resistors ab, ad and ah a current i flows. Thus, I =3i.

    Similarly, in each of the resistors cf, ef and gf a current i will flow towards the junction f. And the current 3i =I will exit out of the arrangement at f.

    Again from symmetry, the current i in ab will get equally divided to flow as i/2 each in bc and be. Again the current i in ad will get equally divided to flow as i/2 each in dc and dg. And the current i in ah will get equally divided to flow as i/2 each in he and hg. We draw a diagram depicting these currents in the corresponding resistors. 
The diagram for Q-44

      Assume that the potential difference across points a and f is V being maintained by cell B. Now apply Kirchhoff's loop law in the loop abcfBa. We go anticlockwise. 

ir + (i/2)r +ir -V =0 

→V =(2i+i/2)r =(5/2)ri 

But we have I = 3i, →i =I/3, hence

V = (5/2)r*(I/3) 

→V =(5r/6)*I.

→V = R*I.

Hence the equivalent resistance R =5r/6.                           

  



 

   45.   Find the equivalent resistances of networks shown in figure (32-E21) between points a and b.  
The figure for Q - 45


ANSWER: (a) Let us name the upper two junctions as c and d. Between junctions c and b, resistors cd and db are in series, thus equivalent to 2r. Now, this 2r is connected in parallel to resistor cb resistance of which is r. Now equivalent resistance r' of 2r and r is given as 

1/r' =1/r +1/2r =3/2r 

→r' =2r/3.  

So the resistances cd, db and cb is replaced with equivalent resistance r' =2r/3 and the given arrangement simplifies to diagram (A) as shown below. 
Diagram for Q-45(a)


   Now ac and cb are in series which is equivalent to

r" =2r/3 +r =5r/3 

    The diagram further simplifies to (B) in which r" =5r/3 and r are in parallel between a and b. The equivalent resistance R between a and b is given as 

1/R = 1/r" +1/r

→1/R = 1/(5r/3) +1/r 

→1/R =3/5r +1/r

→1/R =(3+5)/5r =8/5r 

→R =5r/8.                      


(b) Let us name the resistors and junctions as below. The resistors 1, 2 and 3 are in the parallel connection between points b and O. 
The diagram for Q-45(b)

Hence the equivalent resistance r' of these three resistors will be, 

r' =r/3.

So the given arrangement now simplifies to diagram (B) in which r' and r are connected in series. Hence the equivalent resistance of the given arrangement of resistors between a and b is 

R = r' +r =r/3 +r

→R = 4r/3. 


(c) Let us name the junctions and resistors as below:- 
The diagram for Q-45(c)
Since the resistance of each of the resistors is the same and equal to r, the resistors 1, 2, 3 and 4 in the given arrangement between a and b make a balanced Wheatstone bridge. Hence there will be no current in resistors 5 and 6 and these two will be ineffective.

      The given arrangement now simplifies to diagram (B). The equivalent resistance of resistors 1 and 2 = 2r. Similarly the equivalent resistance of resistors 3 and 4 = 2r. Thus between a and b resistances 2r and 2r are connected in parallel. Hence the equivalent resistance R is given as 

1/R =1/2r +1/2r =1/r

→R = r. 


(d) We can see that each of the resistors r is connected between points a and b. Hence four resistors each of resistance r are connected in parallel between a and b. Thus the equivalent resistance R is given as 

1/R =4/r 

→R =r/4. 


(e) Let us name the resistors and the joints as follows:- 
The diagram for Q-45(e)
On simplification, the given arrangement of resistors in diagram (A) becomes as in diagram (B). In B, the resistors are actually in a balanced Wheatstone bridge. Thus no current in the branch cd and resistor 3 becomes ineffective. 

     Out of four resistors, 2 and 4, as well as 1 and 5, are in series between points a and b. The equivalent resistance of each pair =2r. Finally, between points, a and b two resistors, each of resistance 2r are in parallel combination. The equivalent resistance of this system R is given as,

1/R =1/2r +1/2r

→1/R =1/r 

→R = r.     

               

                 





 

   46.  An infinite ladder is constructed with 1 Ω and 2 Ω resistors as shown in figure (32-E22). (a) Find the effective resistance between points A and B. (b) Find the current that passes through the 2 Ω resistor nearest to the battery.    
The figure for Q-46


ANSWER: (a) Let us assume that the equivalent resistance between the points A and B is R. Since the similar arrangement is infinite, the equivalent resistance between corresponding points in the next loop will also be R. Now the infinite series simplifies to the arrangement shown below.
The diagram for Q-46

  In this arrangement R and 2 Ω resistors are in parallel between points C and D. Hence the equivalent resistance between points C and D =2R/(R+2)

  Thus the equivalent resistance between A and B = 2R/(R+2) +1  

=(2R+R+2)/(R+2)

=(3R+2)/(R+2)

    But we have assumed that the equivalent resistance between points A and B = R. Thus,

 (3R+2)/(R+2) =R     

→3R+2 =R²+2R  

→R² -R -2 =0 

→R² -2R +R -2 =0  --- (i)

→R(R-2) +(R-2) =0 

→(R+1)(R-2) =0 

Either R+1 =0 or R-2 =0. 

Thus either R =-1 or R =2. 

Since we can not take R negative, so R =2 Ω. 

Note: In (i) above, The factorization technique of the quadratic equation has been discussed in my blog post → A simple trick to factorize a quadratic expression in algebra. 


(b) Suppose the current supplied by the battery = i and the current in 2 Ω resistor nearest to the battery =i'. 

     i = 6/R =6/2 A =3 A. 

At point C, the current i divides to i' in 2 Ω resistor and i-i' in R. Since the value of R is also 2 Ω, the current in 2 Ω and R will be equal, i.e. =i/2 =3/2 A =1·5 A. 

 





 

   47.  The emf Ɛ and the internal resistance r of the battery shown in figure (32-E23) are 4.3 V and 1.0 Ω respectively. The external resistance R is 50 Ω. The resistance of the ammeter and the voltmeter are 2·0 Ω and 200 Ω respectively. (a) Find the reading of the two meters. (b) The switch is thrown to the other side. What will be the readings of the two meters now? 
The figure for Q-47


ANSWER: (a) In the iven figure when the switch is left, the resistance of voltmeter and R are in parallel. Their equivalent resistance r',

 1/r' =1/200 +1/50

→1/r' =(1+4)/200

→1/r' =5/200 = 1/40

→r' =40 Ω

Now the resistance of ammeter 2 Ω, r' (=40 Ω) and the internal resistance of battery r =1 Ω are connected in series. The equivalent resistance of the circuit,

R =2 +40 +1 =43 Ω. 

Ɛ = 4·3 V  

Hence the current supplied by the battery, i = Ɛ/R =4·3/43 A

→i = 0·1 A.            

 This will be the current going through the ammeter. Hence the reading of the ammeter = 0·1 A. 

  Suppose a part of this current =i' goes through R and i-i' through the voltmeter. The potential drop across R and the voltmeter will be the same. Hence, 

i'R = (i-i')*200 

→50i' =200i -200i' 

→250i' =200i =200*0·1 

→250i' =20 

→i' = 20/250 A =0·08 A. 

So the potential difference across R or V is = i'*R =0·08*50 = 4 V. So the reading of the voltmeter = 4 V. 


(b) When the switch is thrown to another side, the resistance of the ammeter and R comes in series connection. Their equivalent resistance =50 Ω +2 Ω =52 Ω.

This 52 Ω and the resistance of voltmeter 200 Ω are now in parallel connection. Its equivalent resistance r' is given as  

1/r' =1/52 +1/200

→1/r' =(50+13)/2600  

→1/r' =63/2600

→r' =2600/63 Ω =41·3 Ω

With internal resistance of battery 1 Ω, the total external resistance of the circuit R =41·3 +1 =42·3 Ω.  

Hence the current supplied by the battery, i =Ɛ/R =4·3/42·3 ≈0·1 A.

A part of this current = i' goes through the voltmeter and the rest part =i-i' goes through ammeter+R =r" (say). The potential difference across V and r" will be the same. Hence , 

i'*200 =(i-i')*52

→200i' =52i -52i'

→i' = 52i/252 = 52*0·1/252 

→i' =0·02 A. 

The reading in the ammeter will be i-i' =0·1 -0·02 =0·08 A. 

The reading in the voltmeter will be = i'*200 = 0·02*200 V =4 V.                    





 

   48.  A voltmeter of resistance 400 Ω is used to measure the potential difference across the 100 Ω resistor in the circuit shown in figure (32-E24). (a) What will be the reading of the voltmeter? (b) What was the potential difference across 100 Ω before the voltmeter was connected? 
The figure for Q-48


ANSWER: (a) Given, Ɛ = 84 V. External resistance in the circuit:-

R₁ =100 Ω, R₂ =200 Ω, Resistance of the voltmeter R₃ =400 Ω. 

R₁ and R₃ are connected in parallel, their equivalent resistance R' is given as,

1/R' =1/R₁ +1/R₃ 

→1/R' =1/100 +1/400

→1/R' =5/400 =1/80

→R' =80 Ω.

R' and R₂ are in series connection. Total equivalent external resistance in the circuit, R =R' +R₂ =80+200 =280 Ω.

Hence the current in the circuit, 

i =Ɛ/R =84/280 A =0·3 A

Suppose the voltage drop across R' =V, then V =Ɛ -iR₂

→V =84 -0·3*200 =24 volts.

Hence the reading of the voltmeter will be 24 V. 


(b) Before the connection of the voltmeter, only two resistances R₁ and R₂ are connected in series in the circuit. The equivalent resistance in the circuit,

R =R₁ +R₂ =100+200 =300 Ω 

Hence the current in the circuit, 

i =Ɛ/R =84/300 A

→i =0·28 A  

This current i will go through both the resistors. Hence the potential difference across 100 Ω resistor,

=iR₁ =0·28*100 =28 V. 

  




  

   49.  The voltmeter shown in figure (32-E25) reads 18 V across the 50 Ω resistor. Find the resistance of the voltmeter.  
The figure for Q-49


ANSWER: Since the voltmeter and the 50 Ω resistor are connected in parallel, hence the potential difference across both of them will be equal i.e. =18 V. Let the resistance of the voltmeter = R. 

     The potential difference across 24 Ω will be =30 -18 =12 V. 

Hence the current through 24 Ω,

i =12/24 =0·5 A. 

This current will be divided to go through the 50 Ω resistor and the voltmeter. Since the potential difference across 50 Ω resistor =18 V, the current in 50 Ω resistor will be, i' =18/50 =0·36 A. 

   Thus the current through the voltmeter, i" = i -i' =0·50 -0·36 =0·14 A.

The potential difference across the voltmeter =18 V. 

   Hence the resistance of the voltmeter,

R =18/0·14 ≈ 130 Ω.            

  




 

   50.  A voltmeter consists of a 25 Ω coil connected in series with a 575 Ω resistor. The coil takes 10 mA for full-scale deflection. What maximum potential difference can be measured on this voltmeter?  



ANSWER: At the maximum measured potential difference V' (say), the current in the coil, i =10 mA =0·01 A. 

    The coil (r =25 Ω) is connected in series with a resistor (r' =575 Ω). Hence the same current i =0·01 A will go through it. The equivalent resistance of coil and resistor R = r +r' 

→R = 25 +575 = 600 Ω.

    

Hence the potential difference across R,

=iR =0.01*600 =6 V.  

Thus the voltmeter can measure a maximum potential difference of 6 V.  

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